Question

^{Pinky opened a recurring deposit account of ₹ 300 per month at 12% per annum. If she gets ₹ 8100 at the time of maturity, find the maturity period.}

Answer 1

Given:

Monthly deposit, P = ₹ 300

ate of interest, r = 12% per annum

Maturity Value (MV) = ₹ 8100

MV = `Pxxn + (Pxxn(n+1)xxr)/(2xx12xx100)`

8100 = `300 n + (300 xxn(n+1)xx12)/2400`

`=>8100 = 300n+(3n(n+1))/2`

16200 = 600n + 3n(n + 1)

⇒ 16200 = 600n + 3n² + 3n

⇒ 3n² + 603n − 16200 = 0

n² + 201n − 5400 = 0

`n = (-201+-sqrt(201^2+4xx5400))/2`

`n = (-201+-sqrt(40401+21600))/2`

`n = (-201+-sqrt(62001))/2`

Since 62001 is not a perfect square, we check manually or approximate.

n = 18:

Total deposit: 

= 300 × 18

= ₹ 5400

Interest:

`=(300xx18xx19xx12)/(2xx12xx100)`

`=(300xx342xx12)/(2400)`

`= 1231200/2400`

= ₹ 513

MV = ₹ 5400 + ₹ 513 = ₹ 5913

Let’s try n = 24

Deposit = ₹ 7200

Interest = `(300xx24xx25xx12)/2400`

`=2160000/2400`

= ₹ 900

MV = ₹ 7200 + ₹ 900

= ₹ 8100

Maturity period = 24 months

= 2 years